Question

Helpp. !
What is the product in simplest form? State any restrictions on the variable.

Answer 1

\[\huge \frac{\color{green}{z^2}}{\color{blue}{z+1}}\cdot \frac{\color{orangered}{z^2+3z+2}}{\color{purple}{z^2+3z}}\]

Let's start by factoring out a z from each term in the purple part.\[\huge \frac{\color{green}{z^2}}{\color{blue}{z+1}}\cdot \frac{\color{orangered}{z^2+3z+2}}{\color{purple}{z(z+3)}}\]
From here, we can cancel out a z from the top and bottom,\[\huge \frac{\color{green}{\color{black}{\cancel{\color{green}z}} \cdot z}}{\color{blue}{z+1}}\cdot \frac{\color{orangered}{z^2+3z+2}}{\color{purple}{\color{black}{\cancel{\color{purple}z}}(z+3)}}\]

Answer 2

Next, let's factor the orange part,\[\huge \frac{\color{green}{z}}{\color{blue}{z+1}}\cdot \frac{\color{orangered}{(z+2)(z+1)}}{\color{purple}{(z+3)}}\]

Answer 3

We can make another nice cancellation from here,\[\huge \frac{\color{green}{z}}{\cancel{\color{blue}{z+1}}}\cdot \frac{\color{orangered}{(z+2)}\cancel{\color{orangered}{(z+1)}}}{\color{purple}{(z+3)}}\]

Answer 4

\[\huge \frac{\color{green}{z} \color{orangered}{(z+2)}}{\color{purple}{(z+3)}}\]

Answer 5

Recall that in the land of Math, we're not allowed to divide by 0.

Our denominator will be 0 when \(\large \color{purple}{z=-3}\).
We can't let that happen.

So we'll state that as our restriction on z.\[\large \color{purple}{z \ne -3}\]